Question: 4. The letters "A", "B", "C", "D", "E", and "F" are written on six slips of paper, and the slips are placed into a hat. If the slips are drawn randomly without replacement, what is the probability that "E" is drawn first and "F" is drawn second? 5. The letters "A", "B", "C", "D", "E", and "F" are written on six slips of paper, and the slips are placed into a hat. If the slips are drawn randomly with replacement, what is the probability that "E" is drawn first and "F" is drawn second? 6. A lot of 10 components contains 3 that are defective. Two components are drawn at random and tested. Let A be the event that that the first component drawn is defective, and B be the event that the second component drawn is defective. a. Find P(A). b. Find P(B|A). c. Find P(A and B). d. Are A and B independent? 7. On a TV game show, a contestant is shown 9 products from a grocery store and is asked to choose the three least-expensive items in the set, and then correctly arrange these three items in order of price. In how many ways can the contestant choose the three items?

1. **Problem 4:** Find the probability that "E" is drawn first and "F" is drawn second without replacement from 6 slips labeled A, B, C, D, E, F. 2. **Formula:** For events without replacement, the probability of two sequential events is $$P(E \text{ first and } F \text{ second}) = P(E) \times P(F|E)$$ where $$P(F|E)$$ is the probability of drawing F second given E was drawn first. 3. **Calculations:** - Total slips: $$6$$ - Probability of drawing E first: $$P(E) = \frac{1}{6}$$ - After E is drawn, slips left: $$5$$ - Probability of drawing F second given E first: $$P(F|E) = \frac{1}{5}$$ 4. Multiply probabilities: $$P(E \text{ first and } F \text{ second}) = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30} = 0.0333$$ --- 5. **Problem 5:** Find the probability that "E" is drawn first and "F" is drawn second with replacement from 6 slips. 6. **Formula:** For events with replacement, the draws are independent, so $$P(E \text{ first and } F \text{ second}) = P(E) \times P(F)$$ 7. **Calculations:** - Probability of drawing E first: $$P(E) = \frac{1}{6}$$ - Since replacement, probability of drawing F second: $$P(F) = \frac{1}{6}$$ 8. Multiply probabilities: $$P(E \text{ first and } F \text{ second}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} = 0.0278$$ --- 9. **Problem 6:** A lot of 10 components contains 3 defective. Two components drawn without replacement. Define: - Event A: first component defective - Event B: second component defective 10. **a. Find $$P(A)$$:** - Total components: $$10$$ - Defective: $$3$$ - $$P(A) = \frac{3}{10}$$ 11. **b. Find $$P(B|A)$$:** - After first defective drawn, remaining components: $$9$$ - Remaining defective: $$2$$ - $$P(B|A) = \frac{2}{9}$$ 12. **c. Find $$P(A \text{ and } B)$$:** - $$P(A \text{ and } B) = P(A) \times P(B|A) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15} = 0.0667$$ 13. **d. Are A and B independent?** - Check if $$P(B) = P(B|A)$$ - Calculate $$P(B)$$: - Total defective: 3 - Total components: 10 - $$P(B) = P(B|A)P(A) + P(B|A^c)P(A^c)$$ - $$P(B|A) = \frac{2}{9}$$ - $$P(B|A^c) = \frac{3}{9} = \frac{1}{3}$$ (if first not defective, defective count remains 3 out of 9) - $$P(A) = \frac{3}{10}$$, $$P(A^c) = \frac{7}{10}$$ - Calculate: $$P(B) = \frac{2}{9} \times \frac{3}{10} + \frac{1}{3} \times \frac{7}{10} = \frac{6}{90} + \frac{7}{30} = \frac{1}{15} + \frac{7}{30} = \frac{2}{30} + \frac{7}{30} = \frac{9}{30} = 0.3$$ - Since $$P(B|A) = \frac{2}{9} \approx 0.222 \neq 0.3 = P(B)$$, events A and B are **not independent**. --- 14. **Problem 7:** On a TV game show, contestant chooses 3 least-expensive items from 9 products and arranges them in order. 15. **Formula:** Number of ways to choose and arrange 3 items from 9 is a permutation: $$P(9,3) = \frac{9!}{(9-3)!} = \frac{9!}{6!}$$ 16. **Calculate:** $$9 \times 8 \times 7 = 504$$ 17. **Final answers:** - Problem 4: $$\frac{1}{30} = 0.0333$$ - Problem 5: $$\frac{1}{36} = 0.0278$$ - Problem 6: - a. $$P(A) = \frac{3}{10}$$ - b. $$P(B|A) = \frac{2}{9}$$ - c. $$P(A \text{ and } B) = \frac{1}{15} = 0.0667$$ - d. A and B are **not independent** - Problem 7: Number of ways = $$504$$